Problem:
Square AXYZ is inscribed in equiangular hexagon ABCDEF with X on BC, Y on DE, and Z on EF. Suppose that AB=40 and EF=41(3​−1). What is the side-length of the square?
Answer Choices:
A. 293​
B. 221​2​+241​3​
C. 203​+16
D. 202​+133​
E. 216​ Solution:
Extend EF to H and extend CB to J so that HJ contains A and is perpendicular to lines EF and CB. Let s be the side length of the square and let u=BX. Because ∠ABJ=60∘, it follows that BJ=20 and AJ=203​. Then by the Pythagorean Theorem
AX2=s2=(20+u)2+(203​)2
Because ABCDEF is equiangular, it follows that ED∥AB and so EY∥AB. Also ZY∥AX and thus it follows that ∠EYZ=∠BAX and so △EYZ≅△BAX. Thus EZ=u. Also, ∠HZA=90∘−∠YZE=90∘−∠AXJ=∠JAX; thus △AXJ≅△ZAH and so ZH=203​ and HA=20+u. Moreover, ∠HFA=60∘ and so FH=3​HA​=3​1​(20+u). But EZ+ZH=EF+FH, and so
u+203​=41(3​−1)+3​20+u​
Solving for u yields u=213​−20. Then s2=(213​)2+(203​)2=3⋅292 and therefore s=293​​.
